C++ 中的朗达数

2025年5月24日 | 阅读时长 9 分钟

引言

在数学中，某些特定的数组因其独特的性质而脱颖而出。其中一个分支是Rhonda数，其特点在于其数字和与数字积以一种特定的方式相关联。本文旨在概述什么是Rhonda数，解释其特性，并开发一个简单的 C++ 代码来验证给定的整数是否为Rhonda数。

问题陈述

目标是检查一个数在给定基数b（任何数制，例如十进制数的基数为10）下是否为Rhonda数。

1. 求N的质因数分解

找出N的所有质因数。
计算这些质因数的和。

2. 将质因数之和乘以基数b

计算：b × (质因数之和)

3. 将N转换为b进制并计算其各位数字的乘积

将N转换为其b进制表示形式。
提取各位数字。
计算这些数字的乘积。

4. 检查条件

如果 b × (质因数之和) = (b进制下各位数字的乘积)，则N是基数为b的Rhonda数。
否则，它不是Rhonda数。

示例1：检查1568在基数为10时是否为Rhonda数

让我们验证一下1568在基数为10时是否为Rhonda数。

第1步：求1568的质因数分解。

我们将1568分解为其质因数。

1568 = 2 × 2 × 2 × 2 × 2 × 7 × 7

质因数 2, 2, 2, 2, 2, 7, 7
质因数之和 2+2+2+2+2+7+7=24

第2步：将质因数之和乘以基数(10)

10 × 24 = 240

第3步：将1568转换为10进制并计算各位数字的乘积

1568在10进制下的各位数字 1, 5, 6, 8
各位数字的乘积： 1 × 5 × 6 × 8 = 240

第4步：检查条件

10 × (质因数之和) = 各位数字的乘积

240=240

由于条件成立，1568在基数为10时是Rhonda数。

示例2：检查100在基数为8时是否为Rhonda数

让我们验证一下100在基数为8时是否为Rhonda数。

第1步：求100的质因数分解

100 = 2 × 2 × 5 × 5

质因数2, 2, 5, 5
质因数之和 2+2+5+5=14

第2步：将质因数之和乘以基数(8)

8 × 14 = 112

第3步：将100转换为8进制并计算各位数字的乘积

将100转换为8进制
- 100 ÷ 8 = 12 余 4
- 12 ÷ 8 = 1 余 4
- 1 ÷ 8 = 0 余 1
- 所以，100的8进制表示为144
8进制下的各位数字 1, 4, 4
各位数字的乘积：1 × 4 × 4 = 16

第4步：检查条件

8 × 14 = 112 ≠ 16

由于条件不成立，100在基数为8时不是Rhonda数。

示例 1

现在，让我们编写一个C++程序，用于检查给定数字在基数为10时是否为Rhonda数。

#include <bits/stdc++.h>
using namespace std;

// Function to compute the product of digits in a number
int computeDigitProduct(int num) {
    int product = 1;
    while (num != 0) {
        product *= (num % 10); // Multiply the last digit with the product
        num /= 10; // Remove the last digit
    }
    return product;
}

// Function to compute the sum of all prime factors of a given number
int sumOfPrimeFactors(int num) {
    int sum = 0;
    
    // Count the number of times 2 divides the number
    while (num % 2 == 0) {
        sum += 2;
        num /= 2;
    }
    
    // Check for odd factors from 3 onwards
    for (int factor = 3; factor <= sqrt(num); factor += 2) {
        while (num % factor == 0) {
            sum += factor;
            num /= factor;
        }
    }
    
    // If num is a prime number greater than 2, add it to sum
    if (num > 2)
        sum += num;
    
    return sum;
}

// Function to check if the number is a Rhonda number
bool isRhondaNumber(int number) {
    return 10 * sumOfPrimeFactors(number) == computeDigitProduct(number);
}

// Main function to test the Rhonda number check
int main() {
    cout<<"Enter a number: ";
    int testNumber;
    cin>>testNumber;
    if (isRhondaNumber(testNumber))
        cout << "Yes, it is a Rhonda number.";
    else
        cout << "No, it is not a Rhonda number.";
    
    return 0;
}   

输出 1

Enter a number: 1568
Yes, it is a Rhonda number.

输出 2

Enter a number: 156
No, it is not a Rhonda number.

示例 2

让我们以一个例子来说明C++中的Rhonda数。

#include <bits/stdc++.h>
using namespace std;

// Function to calculate the sum of the prime factors of a number
int sumOfPrimeFactors(int num) {
    int sum = 0;
    int n = abs(num);

    // Handle factor of 2 separately
    while (n % 2 == 0) {
        sum += 2;
        n /= 2;
    }

    // Checking for odd prime factors
    for (int factor = 3; factor * factor <= n; factor += 2) {
        while (n % factor == 0) {
            sum += factor;
            n /= factor;
        }
    }

    // If n is still greater than 2, it must be a prime factor itself
    if (n > 2)
        sum += n;

    return sum;
}

// Function to calculate the product of the digits of a number in a given base
int productOfDigits(int num, int base) {
    if (num == 0) return 0;

    int product = 1;
    int n = abs(num);

    while (n > 0) {
        product *= (n % base);
        n /= base;
    }

    return product;
}

// Function to check if a number is a Rhonda number in a given base
bool isRhonda(int num, int base) {
    return (base * sumOfPrimeFactors(num)) == productOfDigits(num, base);
}

// Function to check if a number is prime
bool isPrime(int num) {
    if (num < 2) return false;
    if (num == 2) return true;
    if (num % 2 == 0) return false;

    for (int i = 3; i * i <= num; i += 2) {
        if (num % i == 0) return false;
    }
    return true;
}

// Function to convert a number to a string representation in a given base
string toBaseString(int num, int base) {
    if (num == 0) return "0";

    string result = "";
    int n = abs(num);

    while (n > 0) {
        int digit = n % base;
        char digitChar = (digit < 10) ? ('0' + digit) : ('a' + digit - 10);
        result = digitChar + result;
        n /= base;
    }

    return result;
}

int main() {
    int maxRhonda = 10;  // Number of Rhonda numbers to find per base
    int maxBase = 16;    // Maximum base to check

    for (int base = 2; base <= maxBase; base++) {
        if (!isPrime(base)) {
            cout << "The first " << maxRhonda << " Rhonda numbers in base " << base << ":\n";

            vector<int> rhondaNumbers;
            int num = 1;

            while (rhondaNumbers.size() < maxRhonda) {
                if (isRhonda(num, base)) {
                    rhondaNumbers.push_back(num);
                }
                num++;
            }

            // Print Rhonda numbers in base 10
            cout << "    In base 10:";
            for (int val : rhondaNumbers) {
                cout << " " << val;
            }
            cout << endl;

            // Print Rhonda numbers in the given base
            if (base != 10) {
                cout << "    In base " << base << ":";
                for (int val : rhondaNumbers) {
                    cout << " " << toBaseString(val, base);
                }
                cout << endl;
            }
        }
    }
    return 0;
}   

输出

The first 10 Rhonda numbers in base 4:
    In base 10: 10206 11935 12150 16031 45030 94185 113022 114415 191149 244713
    In base 4: 2133132 2322133 2331312 3322133 22333212 112333221 123211332 123323233 232222231 323233221
The first 10 Rhonda numbers in base 6:
    In base 10: 855 1029 3813 5577 7040 7304 15104 19136 35350 36992
    In base 6: 3543 4433 25353 41453 52332 53452 153532 224332 431354 443132
The first 10 Rhonda numbers in base 8:
    In base 10: 1836 6318 6622 10530 14500 14739 17655 18550 25398 25956
    In base 8: 3454 14256 14736 24442 34244 34623 42367 44166 61466 62544
The first 10 Rhonda numbers in base 9:
    In base 10: 15540 21054 25331 44360 44660 44733 47652 50560 54944 76857
    In base 9: 23276 31783 37665 66758 67232 67323 72326 76317 83328 126376
The first 10 Rhonda numbers in base 10:
    In base 10: 1568 2835 4752 5265 5439 5664 5824 5832 8526 12985
The first 10 Rhonda numbers in base 12:
    In base 10: 560 800 3993 4425 4602 4888 7315 8296 9315 11849
    In base 12: 3a8 568 2389 2689 27b6 29b4 4297 4974 5483 6a35
The first 10 Rhonda numbers in base 14:
    In base 10: 11475 18655 20565 29631 31725 45387 58404 58667 59950 63945
    In base 14: 4279 6b27 76cd ab27 b7c1 1277d 173da 17547 17bc2 19437
The first 10 Rhonda numbers in base 15:
    In base 10: 2392 2472 11468 15873 17424 18126 19152 20079 24388 30758
    In base 15: a97 aec 35e8 4a83 5269 5586 5a1c 5e39 735d 91a8
The first 10 Rhonda numbers in base 16:
    In base 10: 1000 1134 6776 15912 19624 20043 20355 23946 26296 29070
    In base 16: 3e8 46e 1a78 3e28 4ca8 4e4b 4f83 5d8a 66b8 718e

示例 3

让我们再以一个例子来说明C++中的Rhonda数。

#include <bits/stdc++.h>
using namespace std;

// Function to compute the sum of the prime factors of a number
int computePrimeFactorSum(int num) {
    int factorSum = 0;
    int value = abs(num);

    // Handle factor of 2 separately
    while (value % 2 == 0) {
        factorSum += 2;
        value /= 2;
    }

    // Process odd prime factors
    for (int factor = 3; factor * factor <= value; factor += 2) {
        while (value % factor == 0) {
            factorSum += factor;
            value /= factor;
        }
    }

    // If value is still greater than 2, it's a prime factor
    if (value > 2)
        factorSum += value;

    return factorSum;
}

// Function to calculate the product of digits in a given base
int computeDigitProduct(int num, int base) {
    if (num == 0) return 0;

    int product = 1;
    int value = abs(num);

    while (value > 0) {
        product *= (value % base);
        value /= base;
    }

    return product;
}

// Function to determine if a number is a Rhonda number in a given base
bool checkRhondaNumber(int num, int base) {
    return (base * computePrimeFactorSum(num)) == computeDigitProduct(num, base);
}

// Function to convert a number to a string representation in a specified base
string convertToBaseString(int num, int base) {
    if (num == 0) return "0";

    string result = "";
    int value = abs(num);

    while (value > 0) {
        int digit = value % base;
        char digitChar = (digit < 10) ? ('0' + digit) : ('a' + digit - 10);
        result = digitChar + result;
        value /= base;
    }

    return result;
}

// Function to find and save Rhonda numbers in a given range to a file
void findRhondaNumbersInRange(int base, int startRange, int endRange, const string& filename) {
    ofstream outFile(filename);
    if (!outFile) {
        cerr << "Error opening file!" << endl;
        return;
    }

    outFile << "Rhonda Numbers in base " << base << " from " << startRange << " to " << endRange << ":\n";

    bool found = false;
    for (int num = startRange; num <= endRange; ++num) {
        if (checkRhondaNumber(num, base)) {
            outFile << num << " ";
            found = true;
        }
    }

    if (!found) {
        outFile << "No Rhonda Numbers found in this range.";
    }

    outFile.close();
    cout << "Results saved to " << filename << endl;
}

// Main function to take user input and process Rhonda numbers
int main() {
    int base, startRange, endRange;

    // User input for base
    cout << "Enter the base (e.g., 10 for decimal): ";
    cin >> base;

    // Validate base input
    if (base <= 1) {
        cout << "Error: Base must be greater than 1." << endl;
        return 1;
    }

    // User input for range
    cout << "Enter the start of the range: ";
    cin >> startRange;
    cout << "Enter the end of the range: ";
    cin >> endRange;

    // Validate range input
    if (startRange <= 0 || endRange <= 0 || startRange > endRange) {
        cout << "Error: Invalid range. Please enter positive numbers with start <= end." << endl;
        return 1;
    }

    // Finding Rhonda Numbers and saving results to a file
    string filename = "rhonda_numbers.txt";
    findRhondaNumbersInRange(base, startRange, endRange, filename);

    return 0;
}

输出 1

Enter the base (e.g., 10 for decimal): 10
Enter the start of the range: 1
Enter the end of the range: 3000
Results saved to rhonda_numbers.txt
Note: The file rhonda_numbers.txt contains:
Rhonda Numbers in base 10 from 1 to 3000:
1568 2835

输出 2

Enter the base (e.g., 10 for decimal): 6
Enter the start of the range: 1000
Enter the end of the range: 40000 
Results saved to rhonda_numbers.txt
Note: The file rhonda_numbers.txt contains:
Rhonda Numbers in base 6 from 1000 to 40000:
1029 3813 5577 7040 7304 15104 19136 35350 36992

结论

总之，本文通过创建和编译这个简单的C++代码，解释并详细阐述了或多或少令人惊叹的Rhonda数的特性，该代码可以检查提供的整数是否属于这些数的范畴。该程序最大的优点是其能够在寻找其余Rhonda数时修改基数，从而在编程和数论领域为用户带来激动人心的挑战。

下一个主题C++中的中心四边形数

C++ 中的朗达数

引言

问题陈述

1. 求N的质因数分解

2. 将质因数之和乘以基数b

3. 将N转换为b进制并计算其各位数字的乘积

4. 检查条件

示例1：检查1568在基数为10时是否为Rhonda数

示例 1

示例 2

示例 3

结论

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C++ 中的朗达数

引言

问题陈述

1. 求N的质因数分解

2. 将质因数之和乘以基数b

3. 将N转换为b进制并计算其各位数字的乘积

4. 检查条件

示例1：检查1568在基数为10时是否为Rhonda数

示例 1

示例 2

示例 3

结论

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